Suppose $M$ is an $m$-dimensional manifold and $\gamma \colon \mathbb{R} \rightarrow M$ is a path with $\gamma(0) = p_0$ and $\gamma'(0) = v_0 \in T_{p_0}M$. Let $g \colon \mathbb{R} \rightarrow TM$ denote the path $g(t) = (\gamma(t),\gamma'(t))$. I would like to compute the derivative of $g$ at $0$. Given a chart $\phi \colon U \subset M \rightarrow \mathbb{R}^m$, we get a chart on $TM$ by $$\Phi \colon \pi^{-1}(U) \rightarrow \mathbb{R}^{2m}, ~\Phi(p,v) = d\phi(p,v)= (\phi(p),(d\phi)_pv).$$ The derivative of $g$ is, by definition,$$(dg)_0(\xi) = [d(\Phi \circ g)_0 \xi]_{(p_0,v_0)} = [(~(d\phi)_{p_0}v_0\xi~, ~d\Big((d\phi)_{\gamma(t)} \gamma'(t)\Big)\Big|_{t=0} \xi~)]_{(p_0,v_0)} \in T_{(p_0,v_0)}TM.$$ The expression $(d\phi)_{\gamma(t)}\gamma'(t)$ is a map from $\mathbb{R}$ to $\mathbb{R}^m$, so, theoretically, we understand its derivative. But how can one compute it without knowing an explicit formula for $\phi$? Because I do not know how to differentiate the dependence of $(d\phi)_{\gamma(t)}$. Since the domain $T_{\gamma(t)}M$ of this map is changing in $t$, we need to consider $d\phi$ as the whole map $TM \rightarrow T\mathbb{R}^m = \mathbb{R}^{2m}$. However, doing so yields the following: let $\tilde{\pi}$ be the projection so that $\tilde{\pi}(d\phi(p,v)) = (d\phi)_pv$. Then, if $\zeta \in \mathbb{R}^{2m}$ is such that $[\zeta]_{(p_0,v_0)} = (dg)_0\xi$ in $T_{(p_0,v_0)}TM$, then $$d\Big((d\phi)_{\gamma(t)} \gamma'(t)\Big)\Big|_{t=0} \xi = d\Big(~\tilde{\pi} \circ d\phi \circ g~\Big)\Big|_{t=0}\xi = d\tilde{\pi} \circ d(d\phi))_{(p_0,v_0)} \circ (dg)_0\xi = d\tilde{\pi}(\zeta),$$ which is a circular equation and did not provide me with new insights. Maybe it actually cannot be done without additional info on $\phi$?
Context of this problem: I want to show that for a smooth function $h \colon M \rightarrow \mathbb{R}$, the object $$d^2h_{p_0}(v_0) = \frac{d^2}{dt^2}\Big|_{t=0} (h \circ \gamma)(t),$$ where $\gamma$ is as in the beginning, is a quadratic form (i.e. there is a bilinear form $b \colon T_{p_0} \times T_{p_0} \rightarrow \mathbb{R}$ with $b(v,v) = d^2h_{p_0}(v)$) that is exactly the usual Hessian in local coordinates. So far, I have computed $$d^2h_{p_0}(v_0) = ((d\phi)_{p_0}v_0)^T \cdot \bigg( \frac{\partial^2}{\partial x_i \partial x_j} (h \circ \phi^{-1})(\phi(p_0)) \bigg)_{i,j=1,\dots,m} \cdot \Big(\frac{d^2}{dt^2}\Big|_{t=0} \phi \circ \gamma(t)\Big)$$ and I am stuck computing the term in the right-most bracket, $$\frac{d^2}{dt^2}\Big|_{t=0} \phi \circ \gamma(t) = \frac{d}{dt}\Big|_{t=0} \frac{d}{ds}\Big|_{s=t} \phi \circ \gamma(s) = \frac{d}{dt}\Big|_{t=0} \tilde{\pi} \circ d\phi \circ g = d\tilde{\pi}(\zeta) = \dots$$ I guess this will turn out to be $(d\phi)_{p_0}v_0$...? Cheers!